The force of interaction between two atoms is given by F = \alpha \beta exp ( - x^2 \over \alpha kt ) ; where x is the distance, k is the Boltzmann constant and T is temperature and \alpha and \beta are two constants. The dimension of \beta is :

Units and Measurements PYQ — JEE Main 2019 · 11 Jan · Shift 1 · Q59 with Solution

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Correct answer: B

The interaction force is

F=\alpha\beta\exp\left(-\frac{x^2}{\alpha kT}\right)

where:

We need the dimensions of $\beta$ .

For

\exp\left(-\frac{x^2}{\alpha kT}\right)

the quantity

\frac{x^2}{\alpha kT}

must be dimensionless.

So,

[x^2]=[\alpha kT]

Now,

[x^2]=L^2

Also, $kT$ has dimensions of energy, because Boltzmann constant $k$ has dimensions energy per temperature. Thus,

[kT]=[\text{energy}]=ML^2T^{-2}

Hence,

[\alpha]=\frac{L^2}{ML^2T^{-2}}=M^{-1}T^2

Since exponential is dimensionless,

[F]=[\alpha\beta]

Now,

[F]=MLT^{-2}

Therefore,

[\beta]=\frac{[F]}{[\alpha]}

Substitute $[\alpha]=M^{-1}T^2$ :

[\beta]=\frac{MLT^{-2}}{M^{-1}T^2}

[\beta]=M^2LT^{-4}

[\beta]=M^2LT^{-4}

This corresponds to Option B.

Stored correct answer: B

Our derived answer: B

So, they agree.